Question: Let $h$ be a differentiable function with $h(0)=-1$ and $h'(0)=6$. What is the value of the approximation of $h(0.2)$ using the function's local linear approximation at $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $0.1$ (Choice C) C $0.2$ (Choice D) D $0.3$
Solution: The local linear approximation of $h$ at $x=0$ is achieved using the equation of the line tangent to $h$ at $x=0$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=h'(0)(x-0)+h(0)$. Plugging $h(0)=-1$ and $h'(0)=6$, we obtain $L(x)=6(x-0)-1$. To approximate $h(0.2)$, all we need is to plug $x=0.2$ into $L(x)$. $\begin{aligned} L(0.2)&=6(0.2-0)-1 \\\\ &=6(0.2)-1 \\\\ &=0.2 \end{aligned}$ In conclusion, the approximation of $h(0.2)$ using the function's local linear approximation at $x=0$ is $0.2$.